Also, notice that \(m\) is the sum of six consecutive integers, the smallest of which is \(x\). Therefore, \(m>x\). In other words, \(m\) and \(x\) can't be equal.

(D)
\(n=3x+12\), assuming that \(x\) is the smallest integer. But, the question asks for \(n\) expressed in terms of \(m\), not expressed in terms of \(x\). Eliminate this choice.

(E)Tip: Read the entire question carefully.
If you assume that \(x\) is the smallest of the numbers, and if you solve for the sum of the six consecutive integers instead of \(n\) in terms of \(m\), you may select this wrong answer. A good reason to eliminate this answer immediately is that \(n\) is expressed in terms of \(x\), and not in terms of \(m\).

The integer \(66\) can be expressed as a sum of \(m\) consecutive integers. Which of the following could be the value of \(m\)?

But, \(x=8.5\) is not an integer. We have reached a contradiction. \(66\) cannot be expressed as the sum of six consecutive integers. We can eliminate answer choice (E).

Since the only option that works is II, the answer is (B).

Incorrect Choices:

(A), (C), (D), (E)
See the Solution for why we can eliminate these choices.

Examples on Even and Odd Numbers

Let \(x\) and \(\frac{x+7}{2}\) both be integers. Then \(x\) must be:

(A) a multiple of \(7\)
(B) even
(C) odd
(D) positive
(E) negative

Correct Answer: C

Solution 1:

Tip: Replace variables with numbers.
Try \(x=1\). Then \(\frac{x+7}{2}=\frac{1+7}{2}=\frac{8}{2}=4\), which is an integer. Since \(1\) is not a multiple of \(7\), it is not even, nor is it negative, we can eliminate answers (A), (B), and (E).

Try \(x=2\). Then \(\frac{x+7}{2}=\frac{2+7}{2}=\frac{9}{2}=4.5\), which is not an integer. Therefore, \(x\) can't be even. Eliminate answer (B).

The only remaining option is (C).

Solution 2:

Tip: Look for short-cuts.Tip: Know the properties of even and odd numbers.
The only way \(\frac{x+7}{2}\) can be an integer is if the numerator is divisible by \(2\). That is, the numerator must be even. \(7\) is odd, and only odd \(\pm\) odd \(=\) even. Therefore, \(x\) is odd.

Incorrect Choices:

(A), (B), (D), (E)
See Solution 1 for how we can eliminate these choices by replacing the variable with a numbers.

A
B
C
D
E

Let \(x\) and \(y\) be integers. If \(x + y\) is odd, which of the following is true about \((x+y)^{2}+xy?\)

(A) It is even.
(B) It is odd.
(C) It is even if \(x\) is even.
(D) It is even if \(y\) is even.
(E) It is even if \(xy\) is negative.

If \(x\) and \(y\) are even consecutive integers, \(x<y\), and \(4x - 3y = 10\), what is the value of \(y\)?

Tip: Even numbers: \(\ldots, -6, -4, -2, 0, 2, 4, 6, \ldots , 2n, \ldots\), where \(n\) is an integer.
Consecutive even integers differ from each other by \(2\). Since \(x<y\), \(x+2=y\). We substitute for \(y\) in the given equation:

An evenly-odd number is defined as a positive integer which when once divided evenly by \(2\), cannot be divided evenly by \(2\) again. For example, \(6\) is an evenly-odd number because it can be evenly divided by \(2\) only once: \(6/2=3\).

An evenly-even number is defined as an integer which can be divided evenly by \(2\), the result can itself be divided evenly by \(2\) and so on, until the result is \(1\). For example, \(16\) is an evenly-even number because \(16/2 =8; 8/2=4; 4/2=2; 2/2=1\).

How many evenly-even and evenly-odd integers are there from \(1\) to \(100\)?

By definition, an evenly-odd number, when halved evenly, cannot be halved evenly again. This means that the result of the division by \(2\) is an odd number. So, the evenly-odd numbers are formed when all the odd numbers are multiplied by \(2\). Between \(1\) and \(100\), we count \(25\) evenly-odd numbers:

By definition, evenly-even numbers can be divided repeatedly by \(2\) without a remainder until the result is \(1\). The only numbers that can be repeatedly divided by \(2\) and not leave a remainder are powers of \(2\). We count \(6\) evenly-even numbers between \(1\) and \(100\):

Therefore, there are \(25+6=31\) evenly-odd and evenly-even numbers between \(1\) and \(100\).

Solution 2:

Tip: For an arithmetic sequence: \(a_{n}=a_{1}+(n-1)d\)
Instead of counting the evenly-odd numbers, like we did in Solution 1, we realize that consecutive evenly-odd numbers differ by \(4\), and therefore they form an arithmetic sequence: \(2, 6, 10, \ldots \). We are only interested in those evenly-odd numbers that are between \(1\) and \(100\), so we have the finite sequence \(2, 6, 10 \ldots 94, 98\). To find how many terms are in that sequence, we use the arithmetic sequence rule: \(a_{n}=a_{1}+(n-1)d\), where \(a_{n} = 98\) is the last term of the sequence, \(a_{1=2}\) is the first term in the sequence, \(n\) is the number of terms in the sequence, and \(d=4\) is the difference between two consecutive terms.

So, we have \(25+6=31\) evenly-odd and evenly-even numbers between \(1\) and \(100\).

Incorrect Choices:

(A)
If you only count the evenly-even numbers, instead of both evenly-odd and evenly-even numbers, you will get this wrong answer.

(B)
If you only count the evenly-odd numbers, instead of both evenly-odd and evenly-even numbers, you will get this wrong answer.

(D)
If you count the number of integers from \(1\) to \(49\), instead of the number of odd integers from \(1\) to \(49\) which, when multiplied by \(2\) will form the evenly-odd numbers, then you will get this wrong answer.

(E)
If you count the number of integers from \(1\) to \(49\) and the number of evenly-even numbers from \(1\) to \(100\), you will get this wrong answer.

Review

If you thought these examples difficult and you need to review the material, these links will help: